Taylor Theorem for Two Variables

Consider approximating $z = f(x,y)$

by a quadratic polynomial of $x$

and

. Let

$f(x,y) = a_0 + a_1 x + a_2 y + a_3 x^2 + a_4 xy + a_5 y^2 + R_3$

, where

is an error term. Then find all 2nd order partial derivatives of $z = f(x,y)$

. If

is so small that we can neglect, then

$\displaystyle f(x,y)$	$\displaystyle =$	$\displaystyle a_0 + a_1 x + a_2 y + a_3 x^2 + a_4 xy + a_5 y^2$
$\displaystyle f_{x}(x,y)$	$\displaystyle =$	$\displaystyle a_1 + 2a_3x + a_4y, f_{y}(x,y) = a_2 + a_4 x + 2a_5y$
$\displaystyle f_{xx}(x,y)$	$\displaystyle =$	$\displaystyle a_3, f_{xy}(x,y) = a_4, f_{yy}(x,y) = a_5.$

Now put

. Then

$\displaystyle f(0,0)$	$\displaystyle =$	$\displaystyle a_0, f_x(0,0) = a_1, f_y(0,0) = a_2$
$\displaystyle f_{xx}(0,0)$	$\displaystyle =$	$\displaystyle a_3, f_{xy}(0,0) = a_4, f_{yy}(0,0) = a_5$

Thus we can express all coefficients of $f$

by the 2nd order partial derivatives. Therefore,

$\displaystyle f(x,y)$	$\displaystyle =$	$\displaystyle a_0 + a_1 x + a_2 y + a_3 x^2 + a_4 xy + a_5 y^2$
	$\displaystyle =$	$\displaystyle f(0,0) + f_{x}(0,0)x + f_{y}(0,0)y$
	$\displaystyle +$	$\displaystyle f_{xx}(0,0)x^2 + f_{xy}(0,0)xy + f_{yy}(0,0)y^2$

Theorem 4..7 If is the class $C^{n}$ in the $\delta$ neighborhood of $(x_{0},y_{0})$ , then for $(x_{0} + h,y_{0}+k)$ ,

$\displaystyle f(x_{0} + h,y_{0}+k)$	$\displaystyle =$	$\displaystyle f(x_{0},y_{0}) + \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)f(x_{0},y_{0})$
	$\displaystyle +$	$\displaystyle \frac{1}{2!} \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)^2 f(x_{0},y_{0}) + \cdots$
	$\displaystyle +$	$\displaystyle \frac{1}{(n-1)!} \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)^{n-1} f(x_{0},y_{0}) + R_{n}$

where,

$\displaystyle R_{n} = \frac{1}{n!}\left (h \frac{\partial}{\partial x} + k \fra... ...partial y} \right )^{n} f(x_{0} + \theta h,y_{0} + \theta k) (0 < \theta < 1)$

$\displaystyle \left (h \frac{\partial}{\partial x} + k \frac{\partial}{\partial... ...+ k \frac{\partial}{\partial y} \right )^{m-1} f(x,y), m = 2,3,\cdot \cdot$

NOTE Let $F(t) = f(x_{0} + h t, y_{0} + k t) (0 \leq t \leq 1)$ . Then $F(t)$ is the class $C^{n}$ in $t$ . Thus by Maclausin theorem,

$\displaystyle F(t) = F(0) + \frac{1}{1!}F'(0)t + \frac{1}{2!}F''(0)t^2 + \cdots + \frac{1}{(n-1)!}F^{(n-1)}(0)t^{n-1} + R_{n},$

$\displaystyle R_{n} = \frac{1}{n!}F^{(n)}(\theta t)t^n (0 < \theta < 1).$

Thus for

$\displaystyle F1.$	$\displaystyle =$	$\displaystyle f(x_{0} + h, y_{0} + k) = f(x_{0},y_{0}) + \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)f(x_{0},y_{0})$
	$\displaystyle +$	$\displaystyle \frac{1}{2!} \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)^2 f(x_{0},y_{0}) + \cdots$
	$\displaystyle +$	$\displaystyle \frac{1}{(n-1)!} \left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y} \right)^{n-1} f(x_{0},y_{0}) + R_{n}$

Example 4..17 Given $f(x,y) = e^x \cos{y}$ . Find the Taylor polynomial of 2nd degree at $(0,0)$

SOLUTION We first find all 2nd partial derivatives of $f(x,y) = e^x \cos y$ . $f_{x} = e^x \cos{y}, f_{y} = e^x(-\sin{y}) = -e^x \sin{y}$ , $f_{xx} = e^x \cos{y}$ , $f_{xy} = -e^{x}\sin{y}$ , $f_{yy} = -e^x \cos{y}$ . Theorem4.7, let $x_0 = 0, y_0 = 0, x = h, y = k$

. Then $f(0,0) = 1, f_{x}(0,0) = 1, f_{y}(0,0) = 0, f_{xx}(0,0) = 1, f_{xy}(0,0) = 0, f_{yy}(0,0) = -1$ . Thus

$\displaystyle f(x,y)$	$\displaystyle =$	$\displaystyle f(0,0) + xf_{x}(0,0) + yf_{y}(0,0) + \frac{1}{2!}\left\{x^{2}f_{xx}(0,0)\right.$
	$\displaystyle +$	$\displaystyle \left.2xyf_{xy}(0,0) + y^{2}f_{yy}(0,0) \right\} + R_{3}$
	$\displaystyle =$	$\displaystyle 1 + x + \frac{1}{2!}\left\{x^2 - y^2\right\} + R_{3} \ensuremath{ \blacksquare}$

Exercise 4..17 Find the 2nd degree Taylor polynomial of $f(x,y) = \sin{xy}$ at $(\frac{\pi}{2},1)$ .

SOLUTION $f_{x} = y\cos{xy}, f_{y} = x\cos{xy}, f_{xx} = -y^{2}\sin{xy}, f_{xy} = \cos{xy} -xy\sin{xy}, f_{yy} = -x^{2}\sin{xy}$ . Thus in Theorem4.7, let $x = x_{0} + h = \frac{\pi}{2} + h$ , $y = y_{0} + k = 1 + k$ . Then

$\displaystyle f(x,y)$	$\displaystyle =$	$\displaystyle f(\frac{\pi}{2},1) + (x - \frac{\pi}{2})f_{x}(\frac{\pi}{2},1) + ... ...},1) + \frac{1}{2!}\left\{(x - \frac{\pi}{2})^{2}f_{xx}(\frac{\pi}{2},1)\right.$
	$\displaystyle +$	$\displaystyle \left.2(x - \frac{\pi}{2})(y-1)f_{xy}(\frac{\pi}{2},1) + (y-1)^{2}f_{yy}(\frac{\pi}{2},1) \right\} + R_{3}$
	$\displaystyle =$	$\displaystyle -\frac{1}{2!}\left\{(x - \frac{\pi}{2})^{2} - (x - \frac{\pi}{2})(y-1) - \frac{\pi^{2}}{4}(y-1)^{2}\right\} + R_{3} \ensuremath{ \blacksquare}$

Proof By Taylor theorem, for $(x_{0} + h, y_{0} + k) \in D$ , we have

$\displaystyle f(x_{0} + h,y_{0}+k)$	$\displaystyle =$	$\displaystyle f(x_{0},y_{0}) + hf_{x}(x_{0},y_{0}) + kf_{y}(x_{0} + y_{0})$
	$\displaystyle +$	$\displaystyle \frac{1}{2}[h^2 f_{xx}(x_{0}+\theta h,y_{0}+\theta k) + 2hkf_{xy}(x_{0}+\theta h,y_{0}+\theta k)$
	$\displaystyle +$	$\displaystyle k^2 f_{yy} (x_{0}+\theta h,y_{0}+\theta k)], (0 < \theta < 1)$

Now let $Q = f(x_{0} + h, y_{0} + k) - f(x_{0},y_{0})$ . Then $f_{x}(x_{0},y_{0}) = f_{y}(x_{0},y_{0}) = 0$ and

$\displaystyle Q = \frac{1}{2}[h^2 f_{xx}(x_{0}+\theta h,y_{0}+\theta k) + 2hkf_... ...x_{0}+\theta h,y_{0}+\theta k) + k^2 f_{yy} (x_{0}+\theta h,y_{0}+\theta k) ].$

Next let $A = f_{xx}(x_{0}+\theta h,y_{0}+\theta k), B = f_{xy}(x_{0}+\theta h,y_{0}+\theta k), C = f_{yy} (x_{0}+\theta h,y_{0}+\theta k)$ . Then

$\displaystyle Q = \frac{1}{2}\left(Ah^2 + 2Bhk + Ck^2\right) = \frac{A}{2}\left[(h + \frac{Bk}{A})^2 + \frac{(AC - B^2)k^2}{A^2}\right].$

Thus the sign of $Q$

is determinde by the sign of $A$

and the sign of $AC-B^2$

1. If $\Delta = AC - B^2 > 0, A > 0$ , then since $f(x,y)$ is the class $C^2$ function, for any $h,k$ such that $\vert h\vert,\vert k\vert$ is sufficiently small and never 0 simulteneously, we have $Q > 0$ . Thus, $f(x_{0},y_{0})$ is a local minimum. 2. If $\Delta = AC - B^2 > 0, A < 0$ , then since is the class $C^2$ func, for any $h,k$ such that $\vert h\vert,\vert k\vert$ is sufficiently small and never 0 simulteneously, we have $Q < 0$ . Thus $f(x_{0},y_{0})$ is a local maximum.

3. If $\Delta = AC - B^2 < 0$ and $A > 0$ , then $C < 0$ which gives a saddle point. Similarly, if $\Delta < 0$ and $A < 0$ , then $C > 0$ which give a saddle point. $\blacksquare$

Example 4..18 Find the local extrema of the following function.

$\displaystyle f(x,y) = x^2 + xy + 3y^2 + x +y$

SOLUTION In Example4.16, we found the critical point. Now we check to see whether the function takes a local extremum at the critical point. Now by the 2nd derivative test,

$\displaystyle f_{xx}(x,y) = 2, f_{xy}(x,y) = 1, f_{yy}(x,y) = 6, \Delta = 11$

Thus, $f(-\frac{5}{11}, -\frac{1}{11}) = -\frac{3}{11}$ is a local minimum. $\blacksquare$

Exercise 4..18 Find the local extrema of th following function.

$\displaystyle f(x,y) = x^3 + y^3 - 3xy$

SOLUTION Let $f(x,y) = x^3 + y^3 - 3xy$

. Then we have $f_{x} = 3x^2 - 3y, f_{y} = 3y^2 - 3x$ . If $f(x,y)$

takes the local maxima at $(x_{0},y_{0})$ , then

$\displaystyle f_{x}(x_{0},y_{0})$	$\displaystyle =$	$\displaystyle 3x_{0}^2 - 3y_{0} = 0$	(4.1)
$\displaystyle f_{y}(x_{0},y_{0})$	$\displaystyle =$	$\displaystyle 3y_{0}^2 - 3x_{0} = 0$	(4.2)

Solve this for $x_{0},y_{0}$ . Then substitute the equation $y_{0} = x_{0}^2$ , which is derived from the equation 4.1, to the equation 4.2. Then

$\displaystyle 3x_{0}^4 - 3x_{0} = 3x_{0}(x_{0}^3 - 1) = 0.$

Thus $x_{0} = 0, 1$ . Hence, $(x_{0}=0,y_{0}=0), (x_{0}=1, y_{0} =1)$ .

Now we apply the 2nd derivative test. Since $f_{xx}(x,y) = 6x, f_{xy}(x,y) = -3, f_{yy}(x,y) = 6y$ , at $(0,0)$ we have

$\displaystyle \Delta = f_{xx}(0,0)f_{yy}(0,0) - (f_{xy}(0,0))^2 = -9 < 0.$

This shows that $f(0,0)$

is not extrema. Now at $(1,1)$

, we have

$\displaystyle \Delta$	$\displaystyle =$	$\displaystyle f_{xx}(1,1)f_{yy}(1,1) - (f_{xy}(1,1))^2 = 6\cdot6 - (-3)^2 = 27 > 0$
$\displaystyle A$	$\displaystyle =$	$\displaystyle f_{xx}(1,1) = 6 > 0.$